RMO Part-10!

1)Let A=\(\large(a^{2}+4ab+b^{2})\), where a and b are postive integers.

Prove that A≠2015. (HINT-Use Modular Arithmetic or Try to use the concept of Odd-Even numbers)

2)Prove that any prime number, \(\large(2^{2^n}+1)\) cannot be represented as a difference of 2 fifth powers of integers.(HINT-Expand the expression and You will get something in common)

3)Find all pairs (x,y) where x and y are integers such that \(\large(x^{3}+11^{3}=y^{3})\).(HINT-Try to apply identities)

4)If \(\large(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c})\)

Then Prove that,

\(\large(\frac{1}{a^{3}}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^{3}+b^{3}+c^{3}})\).(HINT-Factorize )

5)Prove that \(\large(2222^{5555}+5555^{2222})\) is divisible by 7.(HINT-Use Modular Arithmetic or Apply the concept of \(a^{n}±b^{n}\).)

Today I have posted some easy problems ,So I'm sure that you will be able to solve all these 5 problems in 1.5 hours.

Also try my Set RMO.

Note by Naitik Sanghavi
3 years, 2 months ago

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Here's my solution to (1).

Assume that ,

\(\large (a+b)^{2}+2ab=2015\)

\(\large (a+b)^{2}=2015-2ab\)

So, \(\large (a+b)^{2}\)=Odd

So,\(\large (a+b)\)=Odd(2q+1)For some positive integer q.

So,\(\large 2ab=4p\)For some integer p.

So, \(\large (2q+1)^{2}+4p=2015\)

\(\large 4(q^{2}+q+p)=2014\)

\(\large q^{2}+q+p=\frac{2014}{4}\)

Now we can see that R.H.S.is not an integer but In L.H.S. all are intgers .Thus it does not satisfy the given conditions.Hence Proved.

Naitik Sanghavi - 3 years, 2 months ago

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4) let \(p(x) = x^3 + lx^2 + mx + p\) with roots a,b,c then \(p = mn\) and \(c = -b\) . Now it easy

Dev Sharma - 3 years, 2 months ago

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Nice!!You got a different to solve this!

Naitik Sanghavi - 3 years, 2 months ago

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5) \(5555=4 \pmod 7\) or \(5555^{2222}=4^{2222} \pmod 7\)

\(4^3=1 \pmod 7\) or \(4^{2220}=1 \pmod 7\) or \(4^{2222}=16=2 \pmod 7\)

Therefore, \(5555^{2222}=2 \pmod 7\).

Similarly,

\(2222=3 \pmod 7\) or \(2222^{5555}=3^{5555} \pmod 7\)

\(3^3=-1 \pmod 7\) or \(3^{5553}=(-1)^{1851}=-1=6 \pmod 7\) or \(3^{5555}=54=5 \pmod 7\)

Therefore, \(5555^{2222}=5 \pmod 7\)

Therefore \(5555^{2222}+2222^{5555}=2+5=7=0 \pmod 7\)

Brilliant Member - 3 years, 2 months ago

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It seems that you have solved all problems of this note,You can my other problems of this set.Try my set Olympiad Preparation Set for an easier level !!

Naitik Sanghavi - 3 years, 2 months ago

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2) Suppose \(2^{2^n}+1=a^5-b^5\),where \(a^5-b^5\) be the difference of two fifth powers.\(a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\ \quad where \quad a>b\).

Since \(2^{2^n}+1 \quad is \quad prime, \quad a-b=1\).

By Fermat's Little theorem, \(2^{2^n}+1=a^5-b^5=a-b \pmod 5=1 \pmod 5\).

Thus \(2^{2^n}+1\) cannot be represented as a difference of two fifth powers.

Moreover, \(2^{2^n}+1\) cannot be represented as the difference of two pth powers, where p is an odd prime.

Brilliant Member - 3 years, 2 months ago

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1) \(a^2 + 4ab + b^2 = 0, 2 or 1 mod4\) but 2015 = 3 mod4. A contradiction.

Dev Sharma - 3 years, 2 months ago

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I seems that you have solved all problems on this Set,You can try my other problems of this set and Try my set Olympiad Preparation setfor an easier level!!

Naitik Sanghavi - 3 years, 2 months ago

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Just a small correction, \(a^2+4ab+b^2\) can also be \(0 \pmod 4\).Try using \pmod for mod. I too did it exactly the same way.

Brilliant Member - 3 years, 2 months ago

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Yes,of course!! You can post your solutions if you have solved any or have an different solution from others!!!

Naitik Sanghavi - 3 years, 2 months ago

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Nice Observation!! I'm having an different solution!!You can try out other problems!!I will post my solutions after some days!!!!!!!

Naitik Sanghavi - 3 years, 2 months ago

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Solution to Q3,

Through Fermat's last theorem

\(a^{ z }+{ b }^{ z }={ c }^{ z }\)

has no natural number solutions for

\(z>2\)

So in,

\({ x }^{ 3 }+{ 11 }^{ 3 }={ y }^{ 3 }\)

Either one of them is 0 or both of them negative,

if \(x\) and \(y\) are negative,

let \(x=-a\) and \(y=-b\) for positive \(a\) and \(b\).

So,

\({ b }^{ 3 }+{ 11 }^{ 3 }={ a }^{ 3 }\)

This has no solutions for positive integers.

So one of \(x\) or \(y\) is \(0\),

if \(x=0\),\(y=11\) and when \(y=0\),\(x=-11\)

Therefore the equation has only two solutions,

\((0,11),(-11,0)\)

Shauryam Akhoury - 2 months, 3 weeks ago

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3) Is (-11,0),(0,11) the only solutions?

Brilliant Member - 3 years, 2 months ago

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Yes!!Correct but you need to prove that there are no other solutions too!!

Naitik Sanghavi - 3 years, 2 months ago

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@naitik sanghavi \(y= \sqrt [3]{x^3+11^3}=\sqrt [3]{(x+11)(x^2-11x+121)}\) which is only a perfect cube when either of the terms are zero. Hence, getting the above solution.

Brilliant Member - 3 years, 2 months ago

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4) This one is quite long! After a lot of solving I got c=-a or c=-b . Substitute both the values and you get the second equation. Am I missing an easier method?

Brilliant Member - 3 years, 2 months ago

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Here's my solution to (4). \(\large \frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\large \frac{a+b}{ab}=\frac{c-(a+b+c)}{(a+b+c)×c}\)

\(\large (a+b)×c×(a+b+c)=-(a+b)×ab\) (Don't remove (a+b) from both sides!Why?See Ahead!!!)

\(\large (a+b)×c×(a+b+c)+(a+b)ab=0\)

\(\large (a+b)[c×(a+c)+b×(a+c)]=0\)

\(\large (a+b)(b+c)(c+a)=0\)

a=-b or b=-c or c=-a.

Now substitute anyone value and it will proved!!

Naitik Sanghavi - 3 years, 2 months ago

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@naitik sanghavi Nice!!.I multiplied (ab+bc+ca) with (a+b+c),did a little bit of factoring and got \(c^2+(a+b)c+ab=0\) and finally solved the quadratic equation to get the above solution.Your solution is much faster upvoted!!

Brilliant Member - 3 years, 2 months ago

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You can also generalize it for powers of n and then by induction it will be proved for 3.

Priyanshu Mishra - 3 years, 2 months ago

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Thanks for your contribution for the Brilliant RMO community :)

Swapnil Das - 3 years, 2 months ago

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Its OK .I am also preparing for RMO so just posting some problems daily..!!!!

Naitik Sanghavi - 3 years, 2 months ago

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